=> x+x+1+x+2+...+x+30 = 1240
=> 31x + (1+2+...+30)=1240
=> 31x + 465 = 1240
=> 31x = 1240 - 465 = 775
=> x =775 : 31 = 25
Vậy x = 25
=> x+x+1+x+2+...+x+30 = 1240
=> 31x + (1+2+...+30)=1240
=> 31x + 465 = 1240
=> 31x = 1240 - 465 = 775
=> x =775 : 31 = 25
Vậy x = 25
Tìm x, biết:
a) \(\left|x-24\right|+\left|y+8\right|=1\)
b)\(\left(x-2\right)^{10}+\left|y-2\right|=0\)
c)\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+30\right)=1240\)
d)\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+2017+2018=2018\)
Giải thích cụ thể giúp mk nha
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
Bài 2 : Tìm x , biết :
a, \(\left(19x+2\cdot5^2\right)\text{ : }14=\left(13-8\right)^2-4^2\)
b, \(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+30\right)=1240\)
c, \(11-\left(-53+x\right)=97\)
d, \(-\left(x+84\right)+213=-16\)
e, \(13-12+11+10-9+8-7-6+5-4+3+2-1=x\)
a) \(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
c) \(|x+7|=20+5.\left(-3\right)\)
Tìm x,biết :
a)\(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
b)x+(x+1)+(x+2)+...+(x+30)=1240
c)11-(-53+x)=97
d)-(x+84)+213=-16
Tìm x biết:
\(a\)) \(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(b\)) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
\(c\)) \(11-\left(-53+x\right)=97\)
Tìm x biết :
\(\left(\frac{x-2}{12}\right)+\left(\frac{x-2}{20}\right)+\left(\frac{x-2}{30}\right)+\left(\frac{x-2}{42}\right)+\left(\frac{x-2}{56}\right)+\left(\frac{x-2}{72}\right)=\frac{16}{9}\)
giúp mk nha
Tìm x biết : \(\frac{\left|x-2\right|}{12}+\frac{\left|x-2\right|}{20}+\frac{\left|x-2\right|}{30}+\frac{\left|x-2\right|}{42}=\frac{70^5}{2^3\cdot21^6}\)
Tìm x, biết:
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\left(x\notin-2;-5;-10;-17\right)\)
\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
Với \(x\notin1;3;8;20\)
\(\frac{x+1}{10}+\frac{2+1}{11}\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x-148}{8}=0\)